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$$\lim_{h\to0}\left\{\frac{h e^{h}(e^{h}+1)+2(1-e^h)e^h}{(1-e^h)^3}\right\}$$

I arrived at this limit when evaluating some residues. Wolfram Alpha tells me it is equal to $-1/6$. But this is such a complicated expression, how would one go about working out the limit? Since each term in the numerator goes to $0$ at the limit, I separated them and tried to use L'Hopital, but this does not work because we get something like $\frac{finite}0-\frac{finite}0$ which is basically $\infty-\infty$. This is not helpful to me.

How can I do it?

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By doing what you did, you have essentially seen that you can't take the limits separately - take them together and you'll get another expression on the numerator which still goes to $0$. Then apply L'Hopital twice more and you're there.

After the first L'hopital application you should get $$\lim_{h\to0}\left\{\frac{(1+h)(e^h+1)+he^h-4e^h+2}{-3(1-e^t)^2}\right\}$$Notice the numerator goes to $2+0-4+2=0$ at the limit, as you want.

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Note that $$\frac{h e^{h}(e^{h}+1)+2(1-e^h)e^h}{(1-e^h)^3}=\frac{h e^{h}(e^{h}+1)+2(1-e^h)e^h}{ h^3 }\frac{h^3}{(1-e^h)^3} =e^h\frac{he^{h}+h+2-2e^h}{ h^3 }\left(\frac{h}{1-e^h}\right)^3\to1\cdot \frac 16\cdot(-1)=-\frac16$$

indeed by l'Hopital

$$\lim_{h\to 0}\frac{he^{h}+h+2-2e^h}{ h^3 }=\lim_{h\to 0}\frac{he^h+e^h+1-2e^h}{3h^2}=\lim_{h\to 0}\frac{he^h+e^h-e^h}{6h}=\lim_{h\to 0}\frac{e^h}{6}=\frac16$$

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  • $\begingroup$ Thanks, this is a nice alternative method. I guess you chose $h^3$ because the Taylor series of $1-e^h$ begins with $h$, similar to Jose's answer right? $\endgroup$ – dialog May 8 '18 at 16:23
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    $\begingroup$ @dialog You are welcome, the choice has been made in order to use standard limit $\frac{h}{1-e^h}\to -1$. $\endgroup$ – user May 8 '18 at 17:01
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The Taylor series of the numerator at $0$ begins with $\frac{h^3}6$, whereas the Taylor series of the denominator at $0$ begins with $-h^3$ (obviously). Therefore, the limit at $0$ is $-\frac{1}6$.

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  • $\begingroup$ Hey good spot, I missed an $e^h$ in the final term. I think that fixes it. $\endgroup$ – dialog May 8 '18 at 12:06
  • $\begingroup$ @dialog I'm glad you agree. Someone thought that I was wrong and downvoted my answer without explaining why. $\endgroup$ – José Carlos Santos May 8 '18 at 12:15
  • $\begingroup$ Yes, I saw this downvote so I upvoted to cancel it. This answer was useful to point out the typo. Also, it revealed that this limit can be done by looking at the Taylor series of the numerator and denominator and comparing - I never thought of this. $\endgroup$ – dialog May 8 '18 at 12:31
  • $\begingroup$ @dialog I've edited my answer, adapting it to the new version of the question. $\endgroup$ – José Carlos Santos May 8 '18 at 12:38
  • $\begingroup$ Thanks a lot, I will remember this Taylor series method in future. $\endgroup$ – dialog May 8 '18 at 16:22

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